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The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a …

I'm interpreting the question as: Measurable, with positive measure, not as "having positive outer measure" (for which the answer is independent of the basic axioms of set theory, as pointed out by Jo …

Ricky: I think I see how to answer the problem under a stronger assumption. Rather than $\mathsf{AD}$, work in $$ {\sf AD}^+ + V=L({\mathcal P}({\mathbb R})). $$ This is a bit unsatisfying, since it …

This is false; take a look at https://en.wikipedia.org/wiki/Analytic_set for a quick introduction. For details, look at Kechris's book on Classical Descriptive Set Theory. There you will find also som …

Consider a subset $\Omega$ of $\mathbb R$ of size $\aleph_1$ and ordered in type $\omega_1$. (This uses the axiom of choice.) Let $\mathcal F$ be the $\sigma$-algebra generated by the initial segmen …

I assume you mean a $\sigma$-additive measure. This is Ulam's measure problem. A positive answer is closely tied up to the existence of real-valued measurable cardinals, so it is equiconsistent with t …

As other answers point out, yes, one needs choice. The popular/natural examples of models of ZF+DC where all sets of reals are measurable are models of determinacy, and Solovay's model. They are relat …

The sets you are calling small are commonly referred to in the literature as "strong measure zero sets." The Borel conjecture is the assertion that any strong measure zero set is countable. This is …

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is that the existence of a nonprincipal ultrafilter d …

No, it is not possible. It is consistent with ZF without choice that the reals are the countable union of countable sets. (*) From this it follows that all sets of reals are Borel. Of course, t …